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aceace Group Title

Logaritm Question!!!

  • 2 years ago
  • 2 years ago

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  1. aceace Group Title
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    Find a relation between x and y that does NOT involve logarithms: log x + log y = log (x+ y)

    • 2 years ago
  2. zepdrix Group Title
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    Can we assume it's a log of base 10?

    • 2 years ago
  3. aceace Group Title
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    yes i think so

    • 2 years ago
  4. AERONIK Group Title
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    logx + log y is never log(x+y) it seems... it is log(xy)

    • 2 years ago
  5. aceace Group Title
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    do you want the answer?

    • 2 years ago
  6. zepdrix Group Title
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    I suppose we could exponentiate both sides.. \[\huge 10^{(\log x + \log y)}=10^{\log(x+y)}\] Using rules of exponents we can split up the left side like this: \[\huge 10^{\log x}*10^{\log y}=10^{\log(x+y)}\] Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". Giving us: \[\huge xy=(x+y)\] \[\huge y=\frac{x}{x-1}\]

    • 2 years ago
  7. zepdrix Group Title
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    Something like that maybe? :o

    • 2 years ago
  8. geoffb Group Title
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    @zepdrix Oooh, sexy!

    • 2 years ago
  9. zepdrix Group Title
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    lol :3

    • 2 years ago
  10. aceace Group Title
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    @zeptr yes thats the answer !!!

    • 2 years ago
  11. zepdrix Group Title
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    Oh cool. :)

    • 2 years ago
  12. zepdrix Group Title
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    Were you just testing us or something? :D lol Knew the answer the whole time huh? XD

    • 2 years ago
  13. zepdrix Group Title
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    @AERONIK , it seems that logx+logy=log(x+y) sometimes. I wouldn't say NEVER. Obviously it doesn't match the rule for logs that we had in mind. But if you plug in y=2, x=2 into the original equation, that should work out ok as one solution :D Since 2+2 is the same as 2*2

    • 2 years ago
  14. aceace Group Title
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    no i had the answers at the back of book but i have no idea how to do it... i still dont get the logic though...

    • 2 years ago
  15. AERONIK Group Title
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    so it is not an identity and just a function you say ?

    • 2 years ago
  16. aceace Group Title
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    yeh just a random equation

    • 2 years ago
  17. zepdrix Group Title
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    Yah it appears it wasn't testing you on whether or not you knew your log identities, they just wanted to see a relationship between x and y. Interesting problem :O

    • 2 years ago
  18. AERONIK Group Title
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    and one more thing in your proof you have assumed the given statement to be correct, can you please give a formal proof for te same?

    • 2 years ago
  19. zepdrix Group Title
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    what now? :o

    • 2 years ago
  20. AERONIK Group Title
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    and is it log to the base 10 or log to the base e ?

    • 2 years ago
  21. zepdrix Group Title
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    For this problem, we have to assume that they are ALL the same base. Otherwise it won't work out the same. If they're a different base, let's say A, then we exponentiate, writing each side with a base A (regardless of what A might be) and it will work out the same! :D

    • 2 years ago
  22. zepdrix Group Title
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    Different than 10 i mean*

    • 2 years ago
  23. AERONIK Group Title
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    oh ok. sorry ..

    • 2 years ago
  24. aceace Group Title
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    10 logx ∗10 logy =10 log(x+y) Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". HOW DID YOU DO THIS STEP... CAN YOU SHOW ME? PLEASE

    • 2 years ago
  25. AERONIK Group Title
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    by defination if y=log x to the base a, then x is nothing but a^y, just use this and you can prove the identity.

    • 2 years ago
  26. zepdrix Group Title
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    \[\large \arcsin( \sin x)=x\]\[\huge e^{\ln x}=x\] Yah this is true of any function and it's inverse :D Remember it in trig with the inverse functions?

    • 2 years ago
  27. aceace Group Title
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    no ...can you do a proof for it please?

    • 2 years ago
  28. zepdrix Group Title
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    |dw:1353225574033:dw| Hmmm this might be a little confusing.. I drew a lot of arrows :| lemme know if it makes some sense.

    • 2 years ago
  29. aceace Group Title
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    i get it ...thanks so much

    • 2 years ago
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